Python equivalent of Haskell's [1..] (to index a list)

Question

I have a list of elements in python. I don't know the number of elements in the list. I would like to add indexes to the list.

In Haskell, I could do the following

zip [1..] "abcdefghijklmnop"
[(1,'a'),(2,'b'),(3,'c'),(4,'d'),(5,'e'),(6,'f'),(7,'g'),(8,'h'),(9,'i'),(10,'j'),(11,'k'),(12,'l'),(13,'m'),(14,'n'),(15,'o'),(16,'p')]

Now imagine that the string was of unknown size. This would still work in Haskell, and the integer list gives as many integers as necessary until the string runs out.

How would one do the equivalent in Python?

I have tried this:

s = "abcdefghijklmnop"
indexedlist = []
for i,c in enumerate(s):
    indexedlist.append((i,c))

>>> indexedlist
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h'), (8, 'i'), (9, 'j'), (10, 'k'), (11, 'l'), (12, 'm'), (13, 'n'), (14, 'o'), (15, 'p')]

And it works, but I'm wondering if there is a shorter/cleaner way, since it is 4 lines of code and feels much.

Answer 1

Just do list(enumerate(s)) . This iterates over the enumerate object and converts it to a list .

Answer 2

您可以通过列表理解来简化它：

>>> [i for i in enumerate(s)]

Answer 3

You can use the range function with zip .

For Python 2:

>>> s = "abcdefghijklmnop"
>>> zip(range(16),s)
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h'), (8, 'i'), (9, 'j'), (10, 'k'), (11, 'l'), (12, 'm'), (13, 'n'), (14, 'o'), (15, 'p')]

For Python 3:

>>> s = "abcdefghijklmnop"
>>> list(zip(range(16),s))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h'), (8, 'i'), (9, 'j'), (10, 'k'), (11, 'l'), (12, 'm'), (13, 'n'), (14, 'o'), (15, 'p')]

Answer 4

Using enumerate is definitely the way to go, but here is a little bit more functional solution with toolz :

from toolz.itertoolz import iterate, zip
zip(iterate(lambda x: x + 1, 0), "abcdefghijklmnop")